\[ \renewcommand{\C}{\mathbb{C}} \renewcommand{\R}{\mathbb{R}} \renewcommand{\Q}{\mathbb{Q}} \]

# 1 Critical points and variation of the gradient

Suppose that \(f(0,0)=0\) and \(\nabla f(0,0)=\begin{pmatrix}0\\0\end{pmatrix}\). We want to review and understand the second derivative test.

## 1.1

We argued that the sign of

\begin{equation*} \nabla f(\delta_x,\delta_y) \cdot \begin{pmatrix}\delta_x\\\delta_y\end{pmatrix} \end{equation*}for all vectors \(\begin{pmatrix}\delta_x\\\delta_y\end{pmatrix}\) close to \(\begin{pmatrix}0\\ 0\end{pmatrix}\) was an efficient way of determinining whether \(f\) has a local max or a local min at \(\begin{pmatrix}0\\0\end{pmatrix}\).

Review why this is the case.

## 1.2

Let us PROVE this fact. Fix a radius \(r\gt 0\). Suppose that

- \(f\) is differentiable on \(D_r(0,0)\)
- \( \nabla f(x,y) \cdot \begin{pmatrix}x\\y\end{pmatrix} \geq 0\) for all \((x,y)\in D_r(0,0)\).

We will PROVE that \(f(x,y)\geq f(0,0)\) for all \((x,y)\in D_r(0,0)\).

Fix \((x,y)\) as above and find a path \(\vec{p}(t)\) that goes from \((0,0)\) to \((x,y)\) with constant velocity as time passes from \(0\) to \(1\). Write the expression for

\begin{equation*} \vec{p}(t) = \begin{pmatrix}p_x(t)\\p_y(t)\end{pmatrix} \end{equation*}## 1.3

Call \(F(t)=f(\vec{p}(t))\). Convince yourself that to prove our statement it is sufficient to show that \(F(1)\geq F(0)\).

## 1.4

Express \(F'(t)\) for any \(t\in[0,1]\) in terms of \(\nabla f(\vec{p}(t))\) and of \(\dot{\vec{p}}(t)\). Also compute \(\dot{\vec{p}}(t)\).

## 1.5

Using the fundamental theorem of calculus write

\begin{equation*} F(1)=F(0)+\int_0^1 F'(t)dt=F(0)= F(0)+\int_0^1 \nabla f(\vec{p}(t)) \cdot \dot{\vec{p}(t)}dt \end{equation*}Show that under our assumptions the integrand is non-negative and deduce the statement of our theorem.

# 2 The Hessian

## 2.1

Review from calculus the second derivative test: if \(F'(t)=0\) and \(F''(t)\gt 0\) then \(F\) has a local min at \(t\).

Why is this true?

## 2.2

We now want to show that the second derivative test in \(2d\) corresponds to checking the second derivative test in \(1d\) over all directions starting from the critical point.

The setup: \(f(x,y)\) is twice differentiable, \(f(0,0)=0\). Fix \(\vec{u}=\begin{pmatrix}u_x,u_y\end{pmatrix}\) a unit vector. Let

\begin{equation*} t\to t\vec{u} \end{equation*}be the parameterization of the line in direction \(\vec{u}\) through \((0,0)\).

## 2.3

Express in terms of \(t\), \(\vec{u}\), and \(f\) the function \(F(t)\) that is the restriction of \(f(x,y)\) to the line mentioned above. Show that \((0,0)\) is critical for \(f\) if and only if \(F'(0)=0\) for however the direction vector \(\vec{u}\) was chosen.

## 2.4

Compute \(F'(t)\) in terms of \(\vec{u}\), \(t\), and \(\nabla f \).

Show that

\begin{equation*} F''(t)= \partial_x^2 f(tu_x,tu_y) u_x^2 + 2\partial_x\partial_yf(tu_x,tu_y) u_xu_y+\partial_y^2 f(tu_x,tu_y) u_y^2 \end{equation*}and in particular show that

\begin{equation*} F''(0)= \partial_x^2 f(0,0) u_x^2 + 2\partial_x\partial_yf(0,0) u_xu_y+\partial_y^2 f(0,0) u_y^2. \end{equation*}See the above recognize the entries as being the coefficients of the Hessian of \(f\) at \((0,0)\)

## 2.5

We now wish to regard the above as an expression that depends on \(u_x\) and \(u_y\) as variables:

\begin{equation*} G(u_x,u_y)= \partial_x^2 f(0,0) u_x^2 + 2\partial_x\partial_yf(0,0) u_xu_y+\partial_y^2 f(0,0) u_y^2. \end{equation*}Why does \(G\) attain max and min over the set of all direction vectors:

\begin{equation*} \vec{u}\in\R^2\qquad \|\vec{u}\|=1? \end{equation*}## 2.6

Parameterize the set \(\vec{u}\in\R^2\qquad \|\vec{u}\|=1\) using polar coordinates. Show that expressing \(H(\theta)=G(cos\theta,\sin\theta)\) as a function of the angle \(\theta\) we have

\begin{equation*} H(\theta)=\cos(\theta)^2 \Big(\partial_x^2 f(0,0) + 2\partial_x\partial_yf(0,0) \tan(\theta)+\partial_y^2 f(0,0) \tan(\theta)^2\Big). \end{equation*}## 2.7

See

\begin{equation*} \Big(\partial_x^2 f(0,0) + 2\partial_x\partial_yf(0,0) \tan(\theta)+\partial_y^2 f(0,0) \tan(\theta)^2\Big). \end{equation*}as a quadratic polynomial in \(\tan(\theta)\). Show that it does not change sign if \(\det(D^2f)\gt 0\) i.e. \(D^2f\) is definite while it DOES change sign if \(\det(D^2f)\lt 0\).

## 2.8

Conclude that you have proved the second derivative test.